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First we take a user input string with the combination of 1’s and 0’s.then create a new string with 1’s,then check if there is any p number of consecutive 1’s is present or not. If present then display FOUND otherwise NOTFOUND.

Binary number ::1111001111 Enter consecutive 1’s :3 Consecutive 1's is Found

Step 1: input a string with the combination of 1’s, it’s stored in the variable X and 0’s and p is the consecutive 1’s in a binary number. Step 2: form a new string of p 1’s. newstring=”1”*p Step 3: check if there is p 1’s at any position. If newstring in X Display “FOUND” Else Display “NOT FOUND” End if

# To check if there is k consecutive 1's in a binary number def binaryno_ones(n,p): # form a new string of k 1's newstr = "1"*p # if there is k 1's at any position if newstr in n: print ("Consecutive 1's is Found") else: print (" Consecutive 1's is Not Found") # driver code n =input("Enter Binary number ::") p = int(input("Enter consecutive 1's ::")) binaryno_ones(n, p)

Enter Binary number ::1111001111 Enter consecutive 1's ::3 Consecutive 1's is Found

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